\(\int \frac {(4+x^3) (-1-x^3+x^4)}{x^8 \sqrt [4]{1+x^3}} \, dx\) [325]

Optimal result

Integrand size = 28, antiderivative size = 28 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=-\frac {4 \left (1+x^3\right )^{3/4} \left (-3-3 x^3+7 x^4\right )}{21 x^7} \]

[Out]

-4/21*(x^3+1)^(3/4)*(7*x^4-3*x^3-3)/x^7

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1849, 1600, 457, 75} \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=-\frac {4 \left (x^3+1\right )^{3/4}}{3 x^3}+\frac {4 \left (x^3+1\right )^{3/4}}{7 x^7}+\frac {4 \left (x^3+1\right )^{3/4}}{7 x^4} \]

[In]

Int[((4 + x^3)*(-1 - x^3 + x^4))/(x^8*(1 + x^3)^(1/4)),x]

[Out]

(4*(1 + x^3)^(3/4))/(7*x^7) + (4*(1 + x^3)^(3/4))/(7*x^4) - (4*(1 + x^3)^(3/4))/(3*x^3)

Rule 75

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1849

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{Pq0 = Coeff[Pq, x, 0]}, Simp[Pq0
*(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] + Dist[1/(2*a*c*(m + 1)), Int[(c*x)^(m + 1)*ExpandToSum
[2*a*(m + 1)*((Pq - Pq0)/x) - 2*b*Pq0*(m + n*(p + 1) + 1)*x^(n - 1), x]*(a + b*x^n)^p, x], x] /; NeQ[Pq0, 0]]
/; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[m, -1] && LeQ[n - 1, Expon[Pq, x]]

Rubi steps \begin{align*} \text {integral}& = \frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}-\frac {1}{14} \int \frac {32 x^2-56 x^3+14 x^5-14 x^6}{x^7 \sqrt [4]{1+x^3}} \, dx \\ & = \frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}-\frac {1}{14} \int \frac {32 x-56 x^2+14 x^4-14 x^5}{x^6 \sqrt [4]{1+x^3}} \, dx \\ & = \frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}-\frac {1}{14} \int \frac {32-56 x+14 x^3-14 x^4}{x^5 \sqrt [4]{1+x^3}} \, dx \\ & = \frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}+\frac {4 \left (1+x^3\right )^{3/4}}{7 x^4}+\frac {1}{112} \int \frac {448+112 x^3}{x^4 \sqrt [4]{1+x^3}} \, dx \\ & = \frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}+\frac {4 \left (1+x^3\right )^{3/4}}{7 x^4}+\frac {1}{336} \text {Subst}\left (\int \frac {448+112 x}{x^2 \sqrt [4]{1+x}} \, dx,x,x^3\right ) \\ & = \frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}+\frac {4 \left (1+x^3\right )^{3/4}}{7 x^4}-\frac {4 \left (1+x^3\right )^{3/4}}{3 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=-\frac {4 \left (1+x^3\right )^{3/4} \left (-3-3 x^3+7 x^4\right )}{21 x^7} \]

[In]

Integrate[((4 + x^3)*(-1 - x^3 + x^4))/(x^8*(1 + x^3)^(1/4)),x]

[Out]

(-4*(1 + x^3)^(3/4)*(-3 - 3*x^3 + 7*x^4))/(21*x^7)

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89

method	result	size
trager	\(-\frac {4 \left (x^{3}+1\right )^{\frac {3}{4}} \left (7 x^{4}-3 x^{3}-3\right )}{21 x^{7}}\)	\(25\)
risch	\(-\frac {4 \left (7 x^{7}-3 x^{6}+7 x^{4}-6 x^{3}-3\right )}{21 x^{7} \left (x^{3}+1\right )^{\frac {1}{4}}}\)	\(35\)
gosper	\(-\frac {4 \left (1+x \right ) \left (x^{2}-x +1\right ) \left (7 x^{4}-3 x^{3}-3\right )}{21 x^{7} \left (x^{3}+1\right )^{\frac {1}{4}}}\)	\(36\)
meijerg	\(\frac {4 \operatorname {hypergeom}\left (\left [-\frac {7}{3}, \frac {1}{4}\right ], \left [-\frac {4}{3}\right ], -x^{3}\right )}{7 x^{7}}+\frac {2 \sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) \left (\frac {5 \pi \sqrt {2}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {9}{4}\right ], \left [2, 3\right ], -x^{3}\right )}{32 \Gamma \left (\frac {3}{4}\right )}-\frac {\left (3-3 \ln \left (2\right )-\frac {\pi }{2}+3 \ln \left (x \right )\right ) \pi \sqrt {2}}{4 \Gamma \left (\frac {3}{4}\right )}-\frac {\pi \sqrt {2}}{\Gamma \left (\frac {3}{4}\right ) x^{3}}\right )}{3 \pi }+\frac {5 \operatorname {hypergeom}\left (\left [-\frac {4}{3}, \frac {1}{4}\right ], \left [-\frac {1}{3}\right ], -x^{3}\right )}{4 x^{4}}+\frac {\sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) \left (-\frac {\pi \sqrt {2}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {5}{4}\right ], \left [2, 2\right ], -x^{3}\right )}{4 \Gamma \left (\frac {3}{4}\right )}+\frac {\left (-3 \ln \left (2\right )-\frac {\pi }{2}+3 \ln \left (x \right )\right ) \pi \sqrt {2}}{\Gamma \left (\frac {3}{4}\right )}\right )}{6 \pi }+\frac {\operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {1}{4}\right ], \left [\frac {2}{3}\right ], -x^{3}\right )}{x}\)	\(180\)

[In]

int((x^3+4)*(x^4-x^3-1)/x^8/(x^3+1)^(1/4),x,method=_RETURNVERBOSE)

[Out]

-4/21*(x^3+1)^(3/4)*(7*x^4-3*x^3-3)/x^7

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=-\frac {4 \, {\left (7 \, x^{4} - 3 \, x^{3} - 3\right )} {\left (x^{3} + 1\right )}^{\frac {3}{4}}}{21 \, x^{7}} \]

[In]

integrate((x^3+4)*(x^4-x^3-1)/x^8/(x^3+1)^(1/4),x, algorithm="fricas")

[Out]

-4/21*(7*x^4 - 3*x^3 - 3)*(x^3 + 1)^(3/4)/x^7

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.51 (sec) , antiderivative size = 177, normalized size of antiderivative = 6.32 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=- \frac {\Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{4} \\ \frac {2}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x \Gamma \left (\frac {2}{3}\right )} - \frac {5 \Gamma \left (- \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, \frac {1}{4} \\ - \frac {1}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x^{4} \Gamma \left (- \frac {1}{3}\right )} - \frac {4 \Gamma \left (- \frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{3}, \frac {1}{4} \\ - \frac {4}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x^{7} \Gamma \left (- \frac {4}{3}\right )} - \frac {\Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x^{\frac {3}{4}} \Gamma \left (\frac {5}{4}\right )} - \frac {4 \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x^{\frac {15}{4}} \Gamma \left (\frac {9}{4}\right )} \]

[In]

integrate((x**3+4)*(x**4-x**3-1)/x**8/(x**3+1)**(1/4),x)

[Out]

-gamma(-1/3)*hyper((-1/3, 1/4), (2/3,), x**3*exp_polar(I*pi))/(3*x*gamma(2/3)) - 5*gamma(-4/3)*hyper((-4/3, 1/
4), (-1/3,), x**3*exp_polar(I*pi))/(3*x**4*gamma(-1/3)) - 4*gamma(-7/3)*hyper((-7/3, 1/4), (-4/3,), x**3*exp_p
olar(I*pi))/(3*x**7*gamma(-4/3)) - gamma(1/4)*hyper((1/4, 1/4), (5/4,), exp_polar(I*pi)/x**3)/(3*x**(3/4)*gamm
a(5/4)) - 4*gamma(5/4)*hyper((1/4, 5/4), (9/4,), exp_polar(I*pi)/x**3)/(3*x**(15/4)*gamma(9/4))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=-\frac {4 \, {\left (7 \, x^{7} - 3 \, x^{6} + 7 \, x^{4} - 6 \, x^{3} - 3\right )}}{21 \, {\left (x^{2} - x + 1\right )}^{\frac {1}{4}} {\left (x + 1\right )}^{\frac {1}{4}} x^{7}} \]

[In]

integrate((x^3+4)*(x^4-x^3-1)/x^8/(x^3+1)^(1/4),x, algorithm="maxima")

[Out]

-4/21*(7*x^7 - 3*x^6 + 7*x^4 - 6*x^3 - 3)/((x^2 - x + 1)^(1/4)*(x + 1)^(1/4)*x^7)

Giac [F]

\[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=\int { \frac {{\left (x^{4} - x^{3} - 1\right )} {\left (x^{3} + 4\right )}}{{\left (x^{3} + 1\right )}^{\frac {1}{4}} x^{8}} \,d x } \]

[In]

integrate((x^3+4)*(x^4-x^3-1)/x^8/(x^3+1)^(1/4),x, algorithm="giac")

[Out]

integrate((x^4 - x^3 - 1)*(x^3 + 4)/((x^3 + 1)^(1/4)*x^8), x)

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=\frac {12\,{\left (x^3+1\right )}^{3/4}+12\,x^3\,{\left (x^3+1\right )}^{3/4}-28\,x^4\,{\left (x^3+1\right )}^{3/4}}{21\,x^7} \]

[In]

int(-((x^3 + 4)*(x^3 - x^4 + 1))/(x^8*(x^3 + 1)^(1/4)),x)

[Out]

(12*(x^3 + 1)^(3/4) + 12*x^3*(x^3 + 1)^(3/4) - 28*x^4*(x^3 + 1)^(3/4))/(21*x^7)